Q17P In two dimensions, show that the... [FREE SOLUTION]

Chapter 1: Q17P (page 18)

In two dimensions, show that the divergence transforms as a scalar under rotations. [Hint: Use Eq. 1.29 to determine $^{v_{y}}$ and $^{v_{z}}$ and the method of Prob. 1.14 to calculate the derivatives. Your aim is to show that
$\frac{鈭� v_{y}}{鈭� y} + \frac{鈭� v_{y}}{鈭� z} = \frac{鈭� v_{y}}{鈭� y} + \frac{鈭� v_{z}}{鈭� z}$

Short Answer

Expert verified

It is shown that $\frac{鈭� v_{y}}{鈭� z} + \frac{鈭� v_{y}}{鈭� y} = \frac{鈭� v_{y}}{鈭� y} + \frac{鈭� v_{y}}{鈭� z}$ . The divergence transforms as a scalar under rotation, in two dimensions as $\frac{鈭� v_{y}}{鈭� z} + \frac{鈭� v_{y}}{鈭� y} = \frac{鈭� v_{y}}{鈭� y} + \frac{鈭� v_{y}}{鈭� z}$

Step by step solution

Define the given information.

It is to be shown that the divergence transforms as a scalar under rotation, in two dimensions as $\frac{鈭� v_{y}}{鈭� z} + \frac{鈭� v_{y}}{鈭� y} = \frac{鈭� v_{y}}{鈭� y} + \frac{鈭� v_{y}}{鈭� z}$

Define a vector.

The vector point function is a function which possess both magnitude and direction, the mathematical representation of a vector point function is as follows:

$a = a_{x} i + a_{y} j + a_{z} k$

Here, $^{ax, ay, xz}$ are the components of vector function $^{a}$ in $^{x, y, z}$ plane respectively.

Set up relation between two set of dimensions.

Write the value of the $^{y}$ and $^{z}$ , as

$y = y \cos 蠒 + z \sin 蠒 . . . . . . . (1) z = - y \sin 蠒 + z \cos 蠒 . . . . . . . (2)$

Multiply on both side of equation (1) as

$z \cos 蠒 = y \cos^{2} 蠒 + z \sin 蠒 \cos 蠒 . . . . . . . . . (3)$

Multiply $^{\sin 鈭�}$ on both side of equation (2) as

role="math" localid="1655961512676" $z \sin 蠒 = - y \sin^{2} 蠒 + z \cos 蠒 \sin 蠒 . . . . . . . . . (4)$

Subtract equation (4) from equation (3)

$y = y \cos 蠒 - z \sin 蠒$

Differentiate above equation with respect to y and z as

$\frac{鈭� y}{鈭� y} = c o s 蠒$

$\frac{鈭� y}{鈭� z} = - \sin 蠒$

Multiply on both side of equation (1) as

$y \sin 蠒 = y cossin 蠒 + z^{2} \sin 蠒 . . . . . (5)$

Multiply $^{\cos 蠒}$ on both side of equation (2) as

$z \cos 蠒 = - y \sin 蠒 \cos 蠒 + z \cos 蠒 . . . . . (6)$

Add equation (5) and equation (6)

$z = y \sin 蠒 + z \cos 蠒$

Differentiate above equation with respect to y and z as

$\frac{鈭� z}{鈭� y} = \sin 蠒 \frac{鈭� z}{鈭� z} = \cos 蠒$

Prove the required equation.

The position of $^{x, y, z}$ axis with respect to $^{x, y, z}$ can be represented via matrix as

$[\frac{v_{y}}{v_{z}}] [\begin{matrix} \cos 蠒 & \sin 蠒 \\ - \sin 蠒 & \cos 蠒 \end{matrix}] [\begin{matrix} v_{y} \\ v_{z} \end{matrix}]$

Expand above matrix as

$\frac{v_{y}}{v_{z}} \begin{matrix} = v_{y} \cos 蠒 + v_{z} \sin 蠒 \\ = - v_{y} \sin 蠒 + v_{z} \cos 蠒 \end{matrix}$

Partially differentiate $^{v_{y}}$ with respect to $^{鈭� y}$ using chain rule, as

role="math" localid="1655964470265" $\frac{鈭� v_{y}}{鈭� y} = \frac{鈭�}{鈭� y} (v_{y} \cos 蠒 + v_{z} \sin 蠒) = \frac{鈭� v_{y}}{鈭� y} \cos 蠒 + \frac{鈭� v_{z}}{鈭� y} \sin 蠒 = (\frac{鈭� v_{y}}{鈭� y} \frac{鈭� y}{鈭� y} + \frac{鈭� v_{y}}{鈭� y} \frac{鈭� z}{鈭� y}) \cos 蠒 + (\frac{鈭� v_{z}}{鈭� y} \frac{鈭� y}{鈭� y} + \frac{鈭� v_{z}}{鈭� z} \frac{鈭� z}{鈭� y}) \sin 蠒$

Substitute $\sin 蠒$ for role="math" localid="1655965188625" $\frac{鈭� z}{鈭� y}, \cos 蠒$ for $\frac{鈭� z}{鈭� z}, \cos 蠒$ for $\frac{鈭� y}{鈭� y}$ and $- \sin 蠒$ for role="math" localid="1655965364288" $\frac{鈭� y}{鈭� z}$ int above expression.

$\frac{鈭� v_{y}}{鈭� y} = (\frac{鈭� v_{y}}{鈭� y} \cos 蠒 + \frac{鈭� v_{y}}{鈭� z} \sin 蠒) \cos 蠒 + (\frac{鈭� v_{z}}{鈭� y} \cos 蠒 + \frac{鈭� v_{z}}{鈭� z} \sin 蠒) \sin 蠒$

Partially differentiate $v_{z}$ with resect to $鈭� z$ ,using chain rule, as

role="math" localid="1655964873819" $\frac{鈭� v_{y}}{鈭� z} = \frac{鈭� v_{y}}{鈭� y} (- v_{y} \sin 蠒 + v_{z} \cos 蠒) = \frac{鈭� v_{y}}{鈭� z} \sin 蠒 + \frac{鈭� v_{z}}{鈭� z} \cos 蠒 = (\frac{鈭� v_{y}}{鈭� y} \frac{鈭� y}{鈭� y z} + \frac{鈭� v_{y}}{鈭� z} \frac{鈭� z}{鈭� z}) \sin 蠒 + (\frac{鈭� v_{z}}{鈭� y} \frac{鈭� y}{鈭� z} + \frac{鈭� v_{z}}{鈭� z} \frac{鈭� z}{鈭� z}) \cos 蠒$

Substitute $\sin 蠒$ for $\frac{鈭� z}{鈭� y}, \cos 蠒$ for $\frac{鈭� z}{鈭� z}, \cos 蠒$ for $\frac{鈭� y}{鈭� y}, \cos 蠒$ and $- \sin 蠒 f o r \frac{鈭� y}{鈭� z}$ for into above expression.

$\frac{鈭� v_{y}}{鈭� z} = - (\frac{鈭� v_{y}}{鈭� y} (- \sin 蠒) + \frac{鈭� v_{y}}{鈭� z} \cos 蠒) \sin 蠒 + (\frac{鈭� v_{z}}{鈭� y} (- \sin 蠒) + \frac{鈭� v_{z}}{鈭� z} \cos 蠒) \cos (蠒)$

Add the expression for $\frac{鈭� v_{y}}{鈭� z} a n d \frac{鈭� v_{y}}{鈭� y} a s, \frac{鈭� v_{y}}{鈭� z} + \frac{鈭� v_{y}}{鈭� y}$

Substitute role="math" localid="1655964045847" $- (\frac{鈭� v_{y}}{鈭� y} (- \sin 蠒) + \frac{鈭� v_{y}}{鈭� z} \cos 蠒) \sin 蠒 + (\frac{鈭� v_{z}}{鈭� y} (- \sin 蠒) + \frac{鈭� v_{z}}{鈭� z} \cos 蠒) \cos (蠒)$ for $\frac{鈭� v_{y}}{鈭� z}$ , and $(\frac{鈭� v_{y}}{鈭� y} \cos 蠒 + \frac{鈭� v_{y}}{鈭� z} \sin 蠒) \cos 蠒 + (\frac{鈭� v_{z}}{鈭� y} (\cos 蠒) + \frac{鈭� v_{z}}{鈭� z} \sin 蠒) \sin 蠒$ for $\frac{鈭� v_{y}}{鈭� y}$ into $\frac{鈭� v_{y}}{鈭� z} + \frac{鈭� v_{y}}{鈭� y}$

role="math" localid="1655963767096" $\frac{鈭� v_{y}}{鈭� z} + \frac{鈭� v_{y}}{鈭� y} = [\begin{matrix} (\frac{鈭� v_{y}}{鈭� y} (- \sin 蠒) + \frac{鈭� v_{y}}{鈭� z} \cos 蠒) \sin 蠒 + (\frac{鈭� v_{z}}{鈭� y} (- \sin 蠒) + \frac{鈭� v_{z}}{鈭� z} \cos 蠒) \cos (蠒) \\ + (\frac{鈭� v_{y}}{鈭� y} \cos 蠒 + \frac{鈭� v_{y}}{鈭� z} \sin 蠒) \cos 蠒 + (\frac{鈭� v_{z}}{鈭� y} (\cos 蠒) + \frac{鈭� v_{z}}{鈭� z} \sin 蠒) \sin 蠒 \end{matrix}] = \frac{鈭� v_{y}}{鈭� y} (\cos^{2} 蠒 + \sin^{2} 蠒) + \frac{鈭� v_{y}}{鈭� z} (\cos^{2} 蠒 + \sin^{2} 蠒) = \frac{鈭� v_{y}}{鈭� y} (1) + \frac{鈭� v_{y}}{鈭� z} (1) = \frac{鈭� v_{y}}{鈭� y} + \frac{鈭� v_{y}}{鈭� z}$

It is obtained that $\frac{鈭� v_{y}}{鈭� z} + \frac{鈭� v_{y}}{鈭� y} = \frac{鈭� v_{y}}{鈭� y} + \frac{鈭� v_{y}}{鈭� z}$ .

Thus, the divergence transforms as a scalar under rotation, in two dimensions as $\frac{鈭� v_{y}}{鈭� z} + \frac{鈭� v_{y}}{鈭� y} = \frac{鈭� v_{y}}{鈭� y} + \frac{鈭� v_{y}}{鈭� z}$

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Short Answer

Step by step solution

Define the given information.

Define a vector.

Set up relation between two set of dimensions.

Prove the required equation.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

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91影视

In two dimensions, show that the divergence transforms as a scalar under rotations. [Hint: Use Eq. 1.29 to determine vyandvzand the method of Prob. 1.14 to calculate the derivatives. Your aim is to show that 鈭�vy鈭�y+鈭�vy鈭�z=鈭�vy鈭�y+鈭�vz鈭�z

Short Answer

Step by step solution

Define the given information.

Define a vector.

Set up relation between two set of dimensions.

Prove the required equation.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Astrophysics

Fluids

Work Energy and Power

Solid State Physics

Fields in Physics

Physics of Motion

Study anywhere. Anytime. Across all devices.